An asteroid is moving directly towards the centre of the earth. When at a distance of 10R (R is the

Q: Why do we neglect the atmosphere in this problem?

We neglect the atmosphere so we can use the principle of conservation of mechanical energy. If we included air resistance, it would act as a non-conservative force, dissipating the asteroid's kinetic energy into heat, and the final speed would be lower.

Q: Why is the initial potential energy negative?

By convention, gravitational potential energy is defined to be zero at an infinite distance. Because gravity is an attractive force, you have to do negative work to bring an object closer from infinity. Therefore, the potential energy is negative everywhere closer than infinity.

Q: Does the mass of the asteroid affect its final speed?

No, it does not. If you look at the energy conservation equation, the mass of the asteroid ($m$) appears in every single term and cancels out completely. Any object, regardless of its mass, would hit the Earth with the same speed under these exact initial conditions.

Q: What if the asteroid was dropped from rest at $10R_e$?

If it was dropped from rest, its initial velocity $v_0$ would be zero. The equation would simplify to $v^2 = \frac{9}{10} v_e^2$. It would hit the Earth with a speed slightly less than the escape velocity.

Q: Why did we use escape velocity in the calculation instead of $G$ and $M_e$?

It's a powerful algebraic shortcut. The problem gives us the numerical value of the escape velocity ($11.2 \text{ km/s}$). By recognizing that $v_e^2 = \frac{2GM_e}{R_e}$, we can substitute this directly into our equation, saving us from having to multiply and divide large astronomical constants.

Question

An asteroid is moving directly towards the centre of the earth. When at a distance of $10R$ ($R$ is the radius of the earth) from the earth's centre, it has a speed of $12 	ext{ km/s}$. Neglecting the effect of earth's atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is $11.2 	ext{ km/s}$)? Give your answer to the nearest integer in km/s ......... [2020, 8 Jan Shift-II]

Sigma Percentile · Accepted Answer

### The Impending Doom

Imagine an asteroid hurtling through the vacuum of space, heading directly towards the center of the Earth. We are given a snapshot of its journey: when it is at a distance of $10R_e$ (where $R_e$ is the radius of the Earth), it is already traveling at a terrifying speed of $12 	ext{ km/s}$.

Our mission is to determine its speed right at the moment of impact. Since we are neglecting the Earth's atmosphere, there is no air resistance to slow it down or burn it up. The only force acting on the asteroid is the conservative force of Earth's gravity.

### The Master Equation: Conservation of Energy

Because gravity is a conservative force, the total mechanical energy of the asteroid remains constant throughout its fall. This means the sum of its kinetic energy ($K$) and gravitational potential energy ($U$) at the initial position will equal the sum at the final position (the Earth's surface).

Let's set up the equation:
$$U_i + K_i = U_f + K_f$$

Substituting the standard formulas for gravitational potential energy ($U = -\frac{GMm}{r}$) and kinetic energy ($K = \frac{1}{2}mv^2$), we get:
$$-\frac{GM_e m}{10R_e} + \frac{1}{2}mv_0^2 = -\frac{GM_e m}{R_e} + \frac{1}{2}mv^2$$

Notice how the mass of the asteroid, $m$, appears in every single term. This is a beautiful feature of gravity: the mass of the falling object cancels out completely. A pebble and a massive asteroid would hit the Earth with the exact same speed under these conditions!

### The Escape Velocity Connection

After canceling $m$, let's rearrange the equation to isolate the final kinetic energy term:
$$\frac{1}{2}v^2 = \frac{1}{2}v_0^2 + \frac{GM_e}{R_e} - \frac{GM_e}{10R_e}$$

Combining the potential energy terms gives:
$$\frac{1}{2}v^2 = \frac{1}{2}v_0^2 + \frac{9GM_e}{10R_e}$$

Now, let's multiply the entire equation by $2$ to solve for $v^2$:
$$v^2 = v_0^2 + \frac{9}{10} \left(\frac{2GM_e}{R_e}ight)$$

Here is where we use a clever algebraic trick. The term $\frac{2GM_e}{R_e}$ is exactly the square of the escape velocity ($v_e$) from the surface of the Earth. The problem graciously provides $v_e = 11.2 	ext{ km/s}$. By substituting $v_e^2$, we avoid dealing with the messy constants $G$, $M_e$, and $R_e$.

$$v^2 = v_0^2 + \frac{9}{10} v_e^2$$

### The Final Impact Speed

Now, it's just a matter of plugging in the numbers. We know $v_0 = 12 	ext{ km/s}$ and $v_e = 11.2 	ext{ km/s}$.

$$v^2 = (12)^2 + 0.9 	imes (11.2)^2$$
$$v^2 = 144 + 0.9 	imes 125.44$$
$$v^2 = 144 + 112.896 = 256.896$$

Taking the square root of this value gives us the final velocity:
$$v = \sqrt{256.896} \approx 16.028 	ext{ km/s}$$

Rounding to the nearest integer, the asteroid will strike the Earth at a blistering speed of **$16 	ext{ km/s}$**.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Conservation of Mechanical Energy$

$Substituting the Values$

$Simplifying the Equation$

$Isolating v^{2}$

$The Escape Velocity Connection$

$Numerical Calculation$

$Final Answer$

$The Way Forward$

The Sigma Insight: Escape Speed and Motion of Satellites

The Impending Doom

The Master Equation

The Escape Velocity Connection

The Final Impact Speed

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Mechanical Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating v2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Escape Velocity Connection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Numerical Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward